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Math Help - Find the minimum value

  1. #1
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    Find the minimum value

    Find the minimum value of (5 cosh x + 3 sinh x).

    How can I solve this?
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  2. #2
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    Quote Originally Posted by geton View Post
    Find the minimum value of (5 cosh x + 3 sinh x).

    How can I solve this?
    Differentiate, put to zero, test nature .... the usual deal.

    Note that to solve when the derivative is equal to zero, you should:

    1. Replace cosh and sinh with their exponential definitions.
    2. Multiply through by 2 e^{-x}.
    3. Solve the resulting quadratic equation in e^x .....
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  3. #3
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    \cosh x = \frac{e^x + e^{-x}}{2}
    \sinh x = \frac{e^x - e^{-x}}{2}

    so, f(x) = 5\cosh x + 3 \sinh x

    f'(x) = 5\sinh x + 3\cosh x = \frac{8e^x - 2e^{-x}}{2} = 4e^x - e^{-x} = \frac{4e^{2x} - 1}{e^x}

    we want to find x such that f'(x) = 0
    since e^x is never zero, we will solve 4e^{2x} - 1 = 0.. solving, we shall obtain

    x = -\ln 2... substitute that x to find the minimum value..

    --------------------------
    to check, f''(x) = f(x) = 5\cosh x + 3 \sinh x = 4e^x + e^{-x} > 0 for any x.
    thus, x = -\ln 2 gives the minimum value for f(x)
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  4. #4
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    Yes I got x = - ln2.

    But its answer is 4.

    I've no idea how can I get this.
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  5. #5
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    Quote Originally Posted by geton View Post
    Yes I got x = - ln2.

    But its answer is 4.

    I've no idea how can I get this.
    Yes, the correct answer is 4.

    f(-\ln 2)=4

    -ln(2) is not the minimum value of 5cosh(x)+3sinh(x), it's just x such that the function is at its minimum.

    Got the difference ?
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  6. #6
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    Quote Originally Posted by Moo View Post
    Yes, the correct answer is 4.

    f(-\ln 2)=4

    -ln(2) is not the minimum value of 5cosh(x)+3sinh(x), it's just x such that the function is at its minimum.

    Got the difference ?

    Yup
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